Question: $f(n) = 3n^{3}+n^{2}+2(h(n))$ $h(n) = -4n$ $ h(f(-2)) = {?} $
First, let's solve for the value of the inner function, $f(-2)$ . Then we'll know what to plug into the outer function. $f(-2) = 3(-2)^{3}+(-2)^{2}+2(h(-2))$ To solve for the value of $f$ , we need to solve for the value of $h(-2)$ $h(-2) = (-4)(-2)$ $h(-2) = 8$ That means $f(-2) = 3(-2)^{3}+(-2)^{2}+(2)(8)$ $f(-2) = -4$ Now we know that $f(-2) = -4$ . Let's solve for $h(f(-2))$ , which is $h(-4)$ $h(-4) = (-4)(-4)$ $h(-4) = 16$